6.5 Images and preimages

If we restrict the codomain of our function to only contain its outputs, we obtain the image of the function.

Definition 6.35 (Image of a function).

Given a function $f:D\to C$ , the set $\{f(x):x\in D\}$ is called the image of $f$ .

Exercise 6.36.

1.

What is the image of the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=3x^{2}-1$ ?
2.

What is the image of the function $g:(0,\infty)\to\mathbb{R}$ defined by $g(x)=3x^{2}-1$ ?

Solution (please try for yourself before looking)

1.

Since $f(x)=3x^{2}-1$ and $x^{2}\geq 0$ for all $x\in\mathbb{R}$ , we have $f(x)\geqslant-1$ for all such $x$ . In particular, for every $y\in[-1,\infty)$ ,

$f\left(\sqrt{\frac{y+1}{3}}\right)=3\left(\sqrt{\frac{y+1}{3}}\right)^{2}-1=y$

and $y\geqslant-1$ implies that $\frac{y+1}{3}\geqslant 0$ so the square root is defined. Hence, the image of $f$ is the set $[-1,\infty)$ .
2.

The function $g$ is the the function $f$ but with a domain $(0,\infty)$ that is a strict subset of the domain of $f$ . This means the image of $g$ should be a subset of the image of $f$ . Indeed, if we take any $y\in[-1,\infty)$ , we know from the previous part that

$f\left(\sqrt{\frac{y+1}{3}}\right)=y,$

but now we need to ensure that $\sqrt{\frac{y+1}{3}}\in(0,\infty)$ . We know that $\sqrt{\frac{y+1}{3}}\geqslant 0$ by definition of the square root, so the only possible issue is if $\sqrt{\frac{y+1}{3}}=0$ and this only occurs when $y=-1$ . Hence, we need to remove $-1$ from the image of $f$ to give us $(-1,\infty)$ for the image of $g$ .

We can also talk about the image of a subset of our domain.

Definition 6.37 (Image of a subset).

Given a function $f:D\to C$ and a subset $S\subseteq D$ , the set

f(S):=\{f(x):x\in S\}

is called the image of $S$ under $f$ .

The image of a subset $S$ of the domain is the set of outputs that our function maps elements of $S$ to. If we instead consider all the inputs that our function maps into a particular subset $U$ of the codomain, we obtain the preimage of that subset $U$ .

Definition 6.38 (Preimage).

Given a function $f:X\to Y$ , the preimage of $U\subseteq Y$ under $f$ is the subset

f^{-1}(U)=\{x\in X:f(x)\in U\}.

Note that the preimage is defined for all functions and all subsets of a function’s codomain. It has nothing to do with the inverse of a function (we haven’t even defined the inverse yet!).

Proposition 6.39.

Let $f:S\rightarrow T$ be a function with $A\subseteq S$ and $B\subseteq T$ . Then

(a)

$A\subseteq f^{-1}(f(A))$ and
(b)

$f(f^{-1}(B))\subseteq B$ .

Proof.

(a)

If we take $x\in A$ , then $f(x)\in f(A)$ . Thus, $x$ is an element of the domain that $f$ maps to an element in $f(A)$ . Hence, $x\in f^{-1}(f(A))$ and so, $A\subseteq f^{-1}(f(A))$ .
(b)

If we take $y\in f(f^{-1}(B))$ , then there exists some $x\in f^{-1}(B)$ such that $f(x)=y$ . But since $x\in f^{-1}(B)$ , we must have $f(x)=y\in B$ . Hence, $f(f^{-1}(B))\subseteq B$ .

Theorem 6.40.

Let $f:X\rightarrow Y$ be a function and $B,B^{\prime}\subseteq Y$ . Then

(a)

$f^{-1}(B\cap B^{\prime})=f^{-1}(B)\cap f^{-1}(B^{\prime})$ and
(b)

$f^{-1}(B\cup B^{\prime})=f^{-1}(B)\cup f^{-1}(B^{\prime})$ .

Proof.

To show the first equality in part (a), note that $x\in f^{-1}(B\cap B^{\prime})$ if and only if $f(x)\in B\cap B^{\prime}$ . Therefore, $f(x)\in B$ and $f(x)\in B^{\prime}$ , which is true if and only if $x\in f^{-1}(B)$ and $x\in f^{-1}(B^{\prime})$ , that is, if and only if $x\in f^{-1}(B)\cap f^{-1}(B^{\prime})$ . Thus, $f^{-1}(B\cap B^{\prime})=f^{-1}(B)\cap f^{-1}(B^{\prime})$ .

The second equation in part (b) has a similar proof to (a) (just change “and” to “or”).